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$$\sum_{n \geq 0} \frac{x^{8n}}{(8n)!}$$

Here's my try:

$$\sum_{n \geq 0, \text{even}} \frac{x^{4n}}{(4n)!}$$

$$=\sum_{n \geq 0} \frac{(-1)^n+1^n}{2} \frac{x^{4n}}{(4n)!}$$

By convergence I can split the sums.

$$=\frac{1}{2} \sum_{n \geq 0} \frac{x^{4n}}{(4n)!}+\frac{1}{2} \sum_{n \geq 0} (-1)^n \frac{x^{4n}}{(4n)!}$$

Now consider,

$$\sum_{n \geq 0} \frac{x^{4n}}{(4n)!}$$

$$=\sum_{n \geq 0, \text{even}} \frac{x^{2n}}{(2n)!}$$

$$=\sum_{n \geq 0} \frac{(-1)^n+1^n}{2} \frac{x^{2n}}{(2n)!}$$

$$=\frac{1}{2} \left( \cos(x)+\cosh (x) \right)$$

Now if I find,

$$\sum_{n \geq 0} (-1)^n \frac{x^{4n}}{(4n)!}$$

I'll be done with the problem. How do I do that?

Bonus question:

Compute

$$\sum_{n \geq 0} \frac{x^{3n}}{(3n)!}$$

For this one I'm out of ideas.

J. M. ain't a mathematician
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Ahmed S. Attaalla
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    I don't have an answer for you, but here is something to think about. As you know, $\sum_{n\geq 0} \frac{x^{2n}}{(2n)!}=\cosh(x)$. This is a function whose second derivative is equal to itself, but whose first derivative is not equal to itself. You can see, similarly, that for any integer $k>0$, $f(x)=\sum_{n \geq 0}\frac{x^{kn}}{(kn)!}$ is a function such that $f'(x), f''(x),\dots,f^{(k-1)}(x) \neq f(x)$, but $f^{(k)}(x)=f(x)$. I remember an undergrad classmate of mine doing a research project on these sorts of functions (over complex numbers). – kccu Mar 20 '17 at 01:18
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    http://www.wolframalpha.com/input/?i=sum(k%3D0,infinity,(-1)%5Ek+x%5E(4k)%2F((4k)!)) – Simply Beautiful Art Mar 20 '17 at 01:22
  • To compute the series, you can just use an integral to approximate the sum see here :https://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula – Zophikel Mar 20 '17 at 01:27
  • You might be interested in the Mittag-Leffler function. – J. M. ain't a mathematician Apr 22 '17 at 17:30

3 Answers3

4

If you allow complex numbers,

$$\sum_{n\ge0}(-1)^n\frac{x^{4n}}{(4n)!}=\sum_{n\ge0}\frac{(xe^{\pi i/4})^{4n}}{(4n)!}=\frac12(\cos(xe^{\pi i/4})+\cosh(xe^{\pi i/4}))$$

By Euler's formula and sum of angles formulas,

$$\Re(\cos(xe^{\pi i/4}))=\Re(\cosh(xe^{\pi i/4}))=\cos\left(\frac x{\sqrt2}\right)\cosh\left(\frac x{\sqrt2}\right)$$

Thus, we may conclude that

$$\sum_{n\ge0}(-1)^n\frac{x^{4n}}{(4n)!}=\cos\left(\frac x{\sqrt2}\right)\cosh\left(\frac x{\sqrt2}\right)$$

As a side note as kccu mentions, we have

$$f(x)=\sum_{n\ge0}\frac{x^{kn}}{(kn)!}\implies f^{(n)}(0)=\begin{cases}1&n\equiv0\mod8\\0&n\not\equiv0\mod8\end{cases}$$

which gives the basic differential equation

$$f(x)=f^{(k)}(x),f(0)=1,f'(0)=f''(0)=0=\dots=,f^{(k-1)}(0)=0$$

Thus, by auxiliary equations, we find that

$$x=x^k\implies x=0,e^{2\pi ni/(k-1)}$$

And then taking real parts and interpolating through the initial derivative values.

Simply Beautiful Art
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3

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{n = 0}^{\infty}{x^{8n} \over \pars{8n}!} & = \sum_{n = 0}^{\infty}{x^{n} \over n!} {\sum_{k = 0}^{7}\bracks{\exp\pars{k\pi\ic/4}}^{n} \over 8} = {1 \over 8}\sum_{k = 0}^{7}\sum_{n = 0}^{\infty}{\bracks{\exp\pars{k\pi\ic/4}x}^{n} \over n!} \\[5mm] & = \bbx{\ds{{1 \over 8}\sum_{k = 0}^{7} \exp\pars{\exp\pars{k\,{\pi \over 4}\,\ic}x}}} \end{align}

Felix Marin
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2

Here is an alternate explicit way to compute the series $\sum_0^{\infty} \frac{x^{pn}}{(pn)!}$ for any positive integer $p$.

Let $\zeta$ be a primitive $p^{th}$ root of unity.

Fix some $n \not\equiv 0 \pmod p$. Then letting $x=\sum_{i=1}^p \zeta^{ni}$ we see that $\zeta^n x = x$ and $\zeta^n \neq 1$, so that $x=0$. When $n \equiv 0 \pmod p$, it is clear that $\sum_{i=1}^p \zeta^{ni}=p$.

Therefore $$\frac{1}{p}\big(e^x + e^{\zeta x} + e^{\zeta^2 x} +...+e^{\zeta^{p-1}x}\big) = \frac{1}{p} \sum_{n=0}^{\infty} (1+\zeta^n+...+\zeta^{(p-1)n})\frac{x^{n}}{n!} = \sum_{n=0}^{\infty} \frac{x^{pn}}{(pn)!}$$

shalop
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